• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16

# Gravitation - Kepler and Newton revision notes and calculations.

Extracts from this document...

Introduction

Gravitation

Kepler (1571-1630) has studied for many years the records of observation on planets and summarised three laws.

Kepler’s Law

1. Each planet moves in an ellipse which has the sun at one focus
2. The line joining the sun and the moving planet sweeps out equal area in equal time
3. the square of the time of revolution of any planet (i.e. T) about the sun is proportional to the cube of the planets’ mean distance from the sun

I.e.   is a constant

Interpretation form Kepler’s laws

1. Kepler’s second law :

The area swept out in a very short time interval (Δt), neglecting the

small triangular region is A

The rate of area swept =

Hence it is a constant.

Compare this equation with the angular momentum

= Constant

This law is in fact an evidence of conservation of angular momentum.

1. Kepler’s third law

About 1666, Newton investigated the motion of the moon, and thought that it was the force of gravity to pull the moon and keep it in its orbit

NB: Time between full moon is 29.5 days but this due to the earth also moving round the sun, the moon is therefore to travel a bit longer

At the earth surface g=9.81m . This is due to the fact that we are nearer the earth centre than the moon (about 1:60)

Middle

m= mg - m

= g -    (< g )

1. At latitude θ

The body describes a circle of radius r centre

Note:

1. Note that  mCosθ is the resultant force of mg and m.

The direction of (or T) is not exactly towards the centre of the earth

except  at the poles and theequator

2) Practical Values

=====================================

Latitude                          / m

9.78

9.79

9.82

9.83

=====================================

Variation of g with height

(i)  r

&

➔                ➔

Let        r = + h

Then

If   h << then

(ii)  r <  (We assume that the earth is a sphere of uniform density)

Consider a small mass placed at A, the spherical shell does not produce

and gravitational field on the point inside it.

&

Since we assume that the earth has uniform density

∝ r

Graphical representation

Practical data

==================================================

Altitude/ h(m)                              /(m)

0                                                  9.81

1000                                                9.80

5                                           8.53

1                                           7.41(Parking Orbit)

3.8                                           0.0027 (Radius of moon orbit)

==================================================

Satellite orbit

The satellite can state in a fixed orbit with a specified radius, r, and

corresponding velocity v by F = ma

--------------- (1)

since

---------------- (2)

(2) →  (1):

Note

For a satellite closed to the earth, stay at a height of about 100 –200km

Then  r

Parking Orbit

The earth is rotating, at this orbit; the satellite can stay constantly

Angular velocity of the earth

F = ma

=

h =

Note:

Conclusion

radius R. Determine the gravitational potential at a point distance r

from centre, where r < R.

Solution:

When  r < R

➔        ➔

When r = R

➔

E.g.7 The masses of the earth and the moon are respectively

and  and they are separated by a

distance

1.  Ignore the motions the earth and the moon , sketch the

gravitational potential V along the line joining them.

1. Find the point where the net gravitational field strength (g) is

zero.

Solution:

1. Choose the earth to be the origin

V = Potential of earth + Potential of moon = +

Near the earth, V is dominated by the field of the earth and near the

moon, V is dominated by the field of the  moon.

E.g.8 Some satellite, with the necessary electronic equipment inside, rises

vertically from the equator when it is fired. At a particular height

the satellite is given a horizontal momentum by firing rocket on its

surface and the satellite then turns into the required orbit.

A satellite is to be put into orbit 500km above the earth’s surface.

If it’s vertical velocity after launching is 2000 at this height.

Calculate the impulse required to put the satellite directly into

Orbit, if its mass is 50kg.    (g = 10   Radius of earth =6400km)

solution:

Suppose u is the velocity required for the orbit, radius R.  Then

Force on the satellite =

u = 7700

Let the impulse P required has the relation shown such that

END

This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Mechanics & Radioactivity essays

1. ## Sir Isaac Newton.

3 star(s)

He therefore proposed and constructed a reflecting telescope. In 1672 Newton was elected a fellow of the Royal Society after donating a reflecting telescope. Also in 1672 Newton published his first scientific paper on light and colour in the Philosophical Transactions of the Royal Society.

2. ## Use of technology in a hospital radiology department. The department of imaging is one ...

observe employment legislations, duty of care to employees, observe employees contracts, provide safe and healthy workplace, giving appropriate training and provide disciplinary procedures. UCLH follows all the HSWA to minimise all unnecessary thing that can harm employee filing and future.

1. ## Investigating the factors affecting tensile strength of human hair.

to a probability level, thus making it possible to identify how likely it is that the values are significantly different. This test is called the Chi squared test. Precautions to ensure reliability * We are assuming ethnic background does not affect our results.

2. ## CIRCULAR MOTION - revision notes and calculations

We must be clear of the cause -result relationship. i.e. we must have centripetal force first, this produces a centripetal acceleration which couples with a tangential velocity to start the circular motion. i.e. Cause Result Centripetal acceleration + imply Circular motion Tangential Velocity (V)

1. ## Investigating the Inverse Square Law

source, and the effective counting space inside the GM tube may not be close to the window, then r = d + d0. * I = intensity * C = corrected count rate - the measured count rate minus the reading for background radiation11 Corrected count rate against 1/(d +

2. ## Measure the earth's gravitational field strength.

The parts equation represent as follows, y = y axis, x = x axis, m = gradient, c = intercept. When this is put into the equation of a straight line the lines are replaced as follows, a = y, sin ??= x, g = m, (-F / m)

1. ## The study of physics has had a large impact on the development of road ...

The car doesn't regain all its original kinetic energy, as some of this is converted to heat and sound energy to reduce damage to passenger area through smaller forces. As crumple zones are placed in strategic locations, the collapse is controlled and energy from impact is directed away from passenger area.

2. ## Charles's Law

motion, and that is why the falling motion of the ball bearing does not need to be considered. A ball bearing will be used, as it is a fairly dense object. This has benefits because although we are ignoring the effect of air resistance in the calculations, we should minimise the affect that it has.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to